For a function f(x), differentiable sufficiently many times, we say that a differentiable polynomial P(D) annihilates f(x) if action of P(D) on f is equal to 0, okay? In this case, we call the differential polynomial P(D) an annihilator of f(x), or sometimes a differential polynomial annihilator of f(x), okay? For example, for any positive integer n, D to the n annihilates x to the k, for all k from 0 to n-1. But that same operator D to the n, it does not annihilates x to the k, for k is greater than or equal to n, okay? So in fact, D to the n annihilates any polynomial of degree strictly less than n, okay? For example, I'd like to find annihilators of the following functions, okay? First, 3 times e to the -x- 2 times x times e to the 2x, okay? How to find this annihilator, okay? Let's look at the term by term then, okay? First, D + 1 e to the -x equals to 0, right? I hope you can see it easily. Again D minus 2 e to the 2x is equals to 0 ,right, okay? I think you know that D minus 2 annihilates e to the 2x, right, okay? Okay, what happen if we multiply e to the 2x by x then? Okay, it will be annihilated by (D-2) square, right? Okay, can you see it, right? So then combining this two, (D+1)e to the -x = 0, and (D-2 )square x e to the 2x =0, okay? So that (D+1) times (D-2) square, okay, it annihilates the whole expression. 3 times e to the negative x- 2 time x e to the 2x, right? So this is an annihilator, okay? This differential polynomial of order 3, this is an annihilator of the given expression, okay? For the second example, -2e to the -x sine 2x, right? You can recognize e to the -x sine of 2x as an imaginary part of exponential -1 plus 2i of x, right, okay? And you also know that, okay, D-(-1 +2i) annihilate exponential (-1+2i)/x, right? Okay, similarly, okay, to find the annihilator of this e to the -x sine 2x, okay, we're seeking quadratic polynomial in D, okay? We are seeking a quadratic polynomial in D, having -1 +- 2i, as its two roots, okay, which is trivially D-(-1+2i) times D-(-1-2i), okay? Expanding this product out, you are going to get D square+2D+5, okay? Then I claim D square+2D+5 annihilates -2 times e to the -x times sine of 2x, right? You can confirm it because, okay, think about that, Linear constant coefficients, homogeneous differential equation. D square+2D+5, and y=0, okay? How to find this general solution, okay? Can you consider this corresponding characteristic equation r squared + 2r + 5 = 0, which has two complex conjugate roots, say -1 +- 2i, okay? And that implies the general solution of this differential equation y = C1 e to the -x cosine of 2x + C2 e to the -x and the sine of 2x, right, okay? And that means what, okay? D squared + 2D + 5 annihilates both e to the -x cosine 2x, and e to the -x and sine 2x, right? So okay, we can claim that this second of the polynomial in D, okay? D squared +2D + 5 is an annihilator of the given function, -2e to the -x and sine 2x, right, okay? As a last example, okay, consider the 3 times e to the -x sine 2x- 2x times e to the -x sine 2x + x square e to the -x cosine 2x, okay. In the example b, we have already seen that, okay, D squared + 2D + 5, okay, annihilates both e to the -x cosine 2x and e to the -x sine 2x, right? Okay, so, okay, this operator, this D square + 2D + 5 annihilates this first part, e to the -x, sine 2x, right? Then what's the annihilator of x times e to the -x sine 2x, right? Okay, we can see that, okay, D square + 2D + 5 square annihilates x times e to the -x and sine 2x, right? By the same token, if we multiply one more x to get x squared times e to the -x and cosine 2x, then last term is annihilated by D squared + 2D + 5, cubed, right? So in fact, D squared + 2D + 5, cubed, annihilates all the other three parts, so it also annihilates the linear combination, okay? So we found that finally D squared + 2D + 5, cubed, is an annihilator of all these expression down here, okay. Generalizing all those examples, we can see rather easily then, for any positive integer n, and for any real numbers alpha and beta, where beta is non-zero, okay. D to the n annihilates x to the k, where k is moving from 0 to n-1. So that means D to the n also now annihilates, any polynomial of degree and most n-1, okay? Then, D-alpha to the n, annihilates, okay, x to the k e to the alpha of x, where k is moving from 0 to n-1, okay? Finally, we can see that D-(alpha + i beta)(D- alpha- i beta), okay? If you look at it, alpha + i beta and alpha- i beta, are complex conjugates. Two numbers, right? So product of these two expression and to the n, okay, it annihilates, okay? Or expanding it out, okay, you will get D square- 2 alpha D + alpha squared + beta squared and to the n, it annihilates, okay. Also such expressions like x to the k, e to the alpha x, cosine beta of x, or x to the k, e to the alpha x, sine beta over x, where k is moving from 0 to n-1, okay? Also at the same operator annihilates any possible linear combination of all these expressions.
